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The formation of hydrogen chloride is exothermic: $$ \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g}) \quad \Delta H=-92.3 \mathrm{~kJ} $$ What are the values of \(\Delta H_{\mathrm{rxn}}\) for $$ \text { (a) } \mathrm{HCl}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g}) $$ (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\)

Short Answer

Expert verified
(a) +92.3 kJ, (b) -184.6 kJ.

Step by step solution

01

Identify Reverse Reaction (a)

For reaction (a): \( \mathrm{HCl}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{Cl}_2(\mathrm{~g}) \), the given reaction is reversed. When a reaction is reversed, the sign of \( \Delta H \) changes. Thus, for the reverse of our original reaction, the enthalpy change, \( \Delta H_{\mathrm{rxn}} \), will be \(+92.3 \, \mathrm{kJ}\).
02

Scale Reaction for Calculation (b)

For reaction (b): \( \mathrm{H}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g}) \), this is essentially the original reaction scaled up by a factor of 2. The enthalpy change \( \Delta H \) for forming \( \mathrm{HCl} \) from \( \frac{1}{2} \mathrm{H}_2 \text{ and } \frac{1}{2} \mathrm{Cl}_2 \) is \(-92.3 \, \mathrm{kJ}\). Therefore, for forming 2 \( \mathrm{HCl} \), \( \Delta H_{\mathrm{rxn}} \) will be twice as much: \(-92.3 \, \mathrm{kJ} \times 2 = -184.6 \, \mathrm{kJ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
In thermochemistry, enthalpy change (\( \Delta H \)) signifies the heat involved in a chemical reaction at constant pressure. This value is crucial to understanding whether a reaction releases or absorbs heat. Enthalpy change can be positive or negative.
  • A positive \( \Delta H \) indicates an endothermic process, where heat is absorbed.
  • A negative \( \Delta H \) signifies an exothermic process, where heat is expelled.

Knowing the sign of \( \Delta H \) allows chemists to predict the energy changes associated with reactions. For example, in the formation of hydrogen chloride, the reaction has a \( \Delta H \) of \(-92.3 \text{ kJ}\), showing it releases heat as an exothermic process. Reversing the reaction or changing the scale will alter \( \Delta H \), as illustrated when doubling the reaction changes \(-92.3 \text{ kJ}\) to \(-184.6 \text{ kJ}\). This knowledge helps us harness chemical reactions effectively, whether in industrial applications or laboratory settings.
Exothermic Reaction
An exothermic reaction is a chemical process that releases energy, usually in the form of heat. This type of reaction makes its surroundings warmer because it expels energy that was stored within the bonds of the reactants.
For example, the formation of hydrogen chloride (HCl) from hydrogen gas (\( \text{H}_2 \)) and chlorine gas (\( \text{Cl}_2 \)) is exothermic. With a \( \Delta H \) of \(-92.3 \text{ kJ}\), the negative sign shows that the reaction releases heat.
To identify whether a reaction is exothermic, look for several key indicators:
  • The \( \Delta H \) value is negative.
  • There is a temperature rise in the surroundings.
  • Heat appears on the product side of a thermochemical equation.
Exothermic reactions are beneficial because they provide energy, making them useful in energy applications such as combusting fuels or as a source of heat in exothermic chemical synthesis.
Chemical Reaction
Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. These processes occur due to changes in the arrangement of electrons around atoms, leading to new compounds.
In the given example, hydrogen chloride forms through an exothermic reaction between hydrogen and chlorine gases:
  • Reactants: \( \frac{1}{2} \text{H}_2 \) and \( \frac{1}{2} \text{Cl}_2 \)
  • Product: \( \text{HCl} \)
This reaction is fascinating because it results from hydrogen and chlorine interacting to create new bonds within HCl molecules. Each phase of a chemical reaction must adhere to the laws of conservation of mass and energy, meaning that the total number of atoms and total energy remains constant, even as the nature of substances changes.
Chemical reactions are manipulated for various purposes, including synthesis of materials, energy production, and biochemical processes in living organisms.

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Most popular questions from this chapter

When lightning strikes, the energy can force atmospheric nitrogen and oxygen to react to make NO: $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H=+181.8 \mathrm{~kJ} $$ (a) Is this reaction endothermic or exothermic? (b) What quantities of reactants and products are $$ \text { assumed if } \Delta H=+181.8 \mathrm{~kJ} \text { ? } $$ (c) What is the enthalpy change when \(3.50 \mathrm{~g}\) nitrogen is reacted with excess \(\mathrm{O}_{2}(\mathrm{~g})\) ?

The law of Dulong and Petit states that the heat capacity of metallic elements is approximately \(25 \mathrm{~J} / \mathrm{mol} \cdot{ }^{\circ} \mathrm{C}\) at \(25^{\circ} \mathrm{C}\). In the 19 th century, scientists used this relationship to obtain approximate atomic masses of metals, from which they determined the formulas of compounds. Once the formula of a compound of the metal with an element of known atomic mass is known, the mass percentage composition of the compound is used to find the atomic mass of the metal. The following example shows the calculations involved. (a) Experimentally, the specific heat of a metal is found to be \(0.24 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). Use the law of Dulong and Petit to calculate the approximate atomic mass of the metal. (b) An oxide of this element is \(6.90 \%\) oxygen by mass. Use the molar mass of \(16.00 \mathrm{~g} / \mathrm{mol}\) for oxygen and the approximate atomic mass found in part (a) to determine the subscripts \(x\) and \(y\) in the formula of the oxide, \(\mathrm{M}_{x} \mathrm{O}_{y} .\) (The mole ratio of the elements you find will not be exactly whole numbers, so considerable rounding is needed to obtain whole numbers in the formula.) (c) From the formula established in part (b), \(x\) mol \(M\) are combined with \(y\) mol \(O .\) Calculate the mass of the metal that is combined with \(y \mathrm{~mol} \mathrm{O}\), using the percent composition of the oxide, and find the atomic mass of the metal. What is the element \(\mathrm{M}\) ?

In the 1880 s, Frederick Trouton noted that the enthalpy of vaporization of \(1 \mathrm{~mol}\) pure liquid is approximately 88 times the boiling point, \(T_{\mathrm{b}}\), of the liquid on the Kelvin scale. This relationship is called Trouton's rule and is represented by the thermochemical equation liquid \(\rightarrow\) gas \(\Delta H=88 \cdot T_{\mathrm{b}}\) joules Combined with an empirical formula from chemical analysis, Trouton's rule can be used to find the molecular formula of a compound, as illustrated here. A compound that contains only carbon and hydrogen is \(85.6 \% \mathrm{C}\) and \(14.4 \% \mathrm{H}\). Its enthalpy of vaporization is \(389 \mathrm{~J} / \mathrm{g}\), and it boils at a temperature of \(322 \mathrm{~K}\). (a) What is the empirical formula of this compound? (b) Use Trouton's rule to calculate the approximate enthalpy of vaporization of one mole of the compound. Combine the enthalpy of vaporization per mole with that same quantity per gram to obtain an approximate molar mass of the compound. (c) Use the results of parts (a) and (b) to find the molecular formula of this compound. Remember that the molecular mass must be exactly a whole-number multiple of the empirical formula mass, so considerable rounding may be needed.

Propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})\), and \(n\) -octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\ell)\), are important components of the widely used fossil fuels. The enthalpy change for combustion of \(1 \mathrm{~mol}\) propane is \(-2219 \mathrm{~kJ},\) and that for \(1 \mathrm{~mol}\) octane is \(-5466 \mathrm{~kJ} .\) Calculate the enthalpy change per gram for each compound.

Use standard enthalpies of formation to calculate the enthalpy change for each of the following reactions at \(298.15 \mathrm{~K}\) and 1 atm. Label each as endothermic or exothermic. (a) The fermentation of glucose to ethyl alcohol and carbon dioxide: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{~s}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+2 \mathrm{CO}_{2}(\mathrm{~g}) $$ (b) The combustion of normal (straight-chain) butane: $$ n-\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{~g})+5 \mathrm{H}_{2} \mathrm{O}(\ell) $$

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